1
00:00:00,000 --> 00:00:02,970
PROFESSOR: This was a
differential equation
2
00:00:02,970 --> 00:00:05,670
for the energy eigenstates phi.
3
00:00:05,670 --> 00:00:08,958
Supposed to be
normalizable functions.
4
00:00:11,770 --> 00:00:15,680
We looked at this equation
and decided we would first
5
00:00:15,680 --> 00:00:16,800
clean out the constant.
6
00:00:16,800 --> 00:00:24,310
We did that by replacing x
by a unit-free coordinate u.
7
00:00:24,310 --> 00:00:28,130
For that we needed a constant
that carries units of length,
8
00:00:28,130 --> 00:00:31,960
and that constant is
given by this combination
9
00:00:31,960 --> 00:00:35,030
of the constants of the problem.
10
00:00:35,030 --> 00:00:37,430
H-bar, m, and omega--
11
00:00:37,430 --> 00:00:40,240
the frequency of the oscillator.
12
00:00:40,240 --> 00:00:44,275
We also defined the
unit-free energy--
13
00:00:44,275 --> 00:00:45,420
calligraphic e.
14
00:00:48,010 --> 00:00:51,330
In terms of which
the real energies
15
00:00:51,330 --> 00:00:55,740
are given by multiples
of h omega over 2.
16
00:00:55,740 --> 00:00:58,120
So the problem has now become--
17
00:00:58,120 --> 00:01:00,850
and this whole
differential equation
18
00:01:00,850 --> 00:01:03,450
turns into this simple
differential equation.
19
00:01:03,450 --> 00:01:07,930
Simple looking--
let's say properly--
20
00:01:07,930 --> 00:01:11,480
differential equation for phi--
21
00:01:11,480 --> 00:01:13,060
as a function of u--
22
00:01:13,060 --> 00:01:16,600
which is the new
rescaled coordinate,
23
00:01:16,600 --> 00:01:19,620
and where the energy
shows up here.
24
00:01:19,620 --> 00:01:21,700
And for some reason,
this equation
25
00:01:21,700 --> 00:01:24,325
doesn't have
normalizable solutions.
26
00:01:24,325 --> 00:01:26,890
Unless those
energies are peculiar
27
00:01:26,890 --> 00:01:34,590
values that allow a
normalizable solution to exist.
28
00:01:34,590 --> 00:01:39,370
We looked at this equation
for u going to infinity,
29
00:01:39,370 --> 00:01:44,280
and realized that e to the
plus minus u squared over 2
30
00:01:44,280 --> 00:01:46,820
are the possible dependencies.
31
00:01:46,820 --> 00:01:49,460
So we said-- without
loss of generality--
32
00:01:49,460 --> 00:01:53,065
that phi could be written
as some function of u
33
00:01:53,065 --> 00:01:56,690
to be determined times
this exponential.
34
00:01:56,690 --> 00:02:03,730
And we hope for a function
that may be a polynomial.
35
00:02:03,730 --> 00:02:06,790
So that the dependence
at the infinity
36
00:02:06,790 --> 00:02:09,620
is governed by this factor.
37
00:02:09,620 --> 00:02:13,930
So with this for
the function phi,
38
00:02:13,930 --> 00:02:18,170
we substitute back into
the differential equation.
39
00:02:18,170 --> 00:02:20,230
Now the unknown is h.
40
00:02:20,230 --> 00:02:24,400
So you can take the derivatives
and find this differential
41
00:02:24,400 --> 00:02:28,600
equation for h-- a second
order differential equation.
42
00:02:28,600 --> 00:02:30,300
And that's the equation.
43
00:02:30,300 --> 00:02:32,860
It may look a little
more complicated
44
00:02:32,860 --> 00:02:34,840
than the equation
we started with,
45
00:02:34,840 --> 00:02:36,465
but it's much simpler, actually.
46
00:02:39,220 --> 00:02:42,340
There would be no polynomial
solution of this equation,
47
00:02:42,340 --> 00:02:44,364
but there may be a
polynomial solution
48
00:02:44,364 --> 00:02:46,886
of the second equation.
49
00:02:46,886 --> 00:02:49,840
So we have to solve
this equation now.
50
00:02:49,840 --> 00:02:55,610
And the way to do it is to
attempt a serious expansion.
51
00:02:55,610 --> 00:03:03,352
So we would try to write h
of u equals the sum over j.
52
00:03:03,352 --> 00:03:05,540
Equals 0 to infinity.
53
00:03:09,540 --> 00:03:12,340
Ak, u to the k.
54
00:03:15,100 --> 00:03:18,630
P equals 0 to infinity.
55
00:03:18,630 --> 00:03:21,750
Now, one way to
proceed with this
56
00:03:21,750 --> 00:03:27,310
is to plot this expansion into
the differential equation.
57
00:03:27,310 --> 00:03:28,830
You will get three sums.
58
00:03:28,830 --> 00:03:31,570
You will have to shift indices.
59
00:03:31,570 --> 00:03:35,350
It's kind of a
little complicated.
60
00:03:35,350 --> 00:03:39,590
Actually, there's a simpler
way to do this in which you
61
00:03:39,590 --> 00:03:42,830
think in the following way.
62
00:03:42,830 --> 00:03:45,355
You have this series
and you imagine
63
00:03:45,355 --> 00:03:53,933
there's a term aj, u to the j,
plus aj plus 1, u to the j plus
64
00:03:53,933 --> 00:03:57,330
1, plus aj plus 2.
65
00:03:57,330 --> 00:03:59,028
U to the j plus 2.
66
00:04:02,772 --> 00:04:16,690
And you say, let me
look at the terms with u
67
00:04:16,690 --> 00:04:23,360
to the j in the
differential equation.
68
00:04:23,360 --> 00:04:29,180
So just look at the terms
that have a u to the power j.
69
00:04:29,180 --> 00:04:32,330
So from this second--
70
00:04:32,330 --> 00:04:37,470
h vu squared-- what do we get?
71
00:04:37,470 --> 00:04:42,210
Well, to get a term that has
a u to the j, you must start--
72
00:04:42,210 --> 00:04:46,040
if you take two derivatives
and to end up with u to the j--
73
00:04:46,040 --> 00:04:49,270
you must have started with this.
74
00:04:49,270 --> 00:04:50,910
U to the j plus 2.
75
00:04:50,910 --> 00:04:57,360
So this gives you j
plus 2, j plus 1--
76
00:04:57,360 --> 00:04:59,240
taking the two derivatives--
77
00:04:59,240 --> 00:05:04,070
aj plus 2, u to the j.
78
00:05:04,070 --> 00:05:11,790
From the series, the term with
u to the j from the second hvu
79
00:05:11,790 --> 00:05:12,820
squared is this one.
80
00:05:15,410 --> 00:05:21,280
How about for minus 2u dh, du?
81
00:05:25,500 --> 00:05:34,110
Well, if I start with h and
differentiate and then multiply
82
00:05:34,110 --> 00:05:39,960
by u, I'm going to get u to
the j starting from u to the j.
83
00:05:39,960 --> 00:05:43,740
Because when I differentiate
I'll get u to the j minus 1,
84
00:05:43,740 --> 00:05:46,200
but the u will bring it back.
85
00:05:46,200 --> 00:05:52,320
So this time I get minus 2 a j--
86
00:05:52,320 --> 00:05:54,345
or minus 2.
87
00:05:54,345 --> 00:05:57,201
One derivative j.
88
00:05:57,201 --> 00:06:00,500
That's aj, u to the j.
89
00:06:03,860 --> 00:06:09,202
So it's from this
[? step. ?] Minus 2.
90
00:06:09,202 --> 00:06:11,730
You differentiate
and you get that.
91
00:06:11,730 --> 00:06:18,750
From the last term,
e minus 1 times h,
92
00:06:18,750 --> 00:06:27,360
it's clearly e minus
1 times aj u to the j.
93
00:06:27,360 --> 00:06:33,930
So these are my three terms that
we get from the differential
94
00:06:33,930 --> 00:06:34,570
equation.
95
00:06:34,570 --> 00:06:40,270
So at the end of the
day, what have we gotten?
96
00:06:40,270 --> 00:06:54,030
We've gotten j plus 2, j plus
1, aj plus 2, minus 2jaj,
97
00:06:54,030 --> 00:06:58,770
plus e minus 1 aj.
98
00:06:58,770 --> 00:07:03,390
All multiplied by u to the j.
99
00:07:03,390 --> 00:07:06,752
And that's what we
get for u to the j.
100
00:07:06,752 --> 00:07:10,760
So if you wish, for the
whole differential equation--
101
00:07:13,650 --> 00:07:16,790
all of the
differential equation--
102
00:07:16,790 --> 00:07:24,710
you get the sum from j equals
0 to infinity of these things.
103
00:07:24,710 --> 00:07:28,400
And that should
be equal to zero.
104
00:07:28,400 --> 00:07:30,560
So this is the
whole left-hand side
105
00:07:30,560 --> 00:07:31,775
of the differential equation.
106
00:07:35,090 --> 00:07:38,090
We calculated what is
the term u to the j.
107
00:07:38,090 --> 00:07:42,620
And there will be terms
from u to the zeroth
108
00:07:42,620 --> 00:07:45,100
to u to the infinity.
109
00:07:45,100 --> 00:07:47,390
So that's the whole thing.
110
00:07:47,390 --> 00:07:51,350
And we need this differential
equation to be solved.
111
00:07:51,350 --> 00:07:53,150
So this must be zero.
112
00:07:53,150 --> 00:07:57,440
And whenever you have a function
of u like a polynomial-- well,
113
00:07:57,440 --> 00:08:01,040
we don't know if it's a
polynomial-- and it stops.
114
00:08:01,040 --> 00:08:03,560
But if you have a
function of u like this,
115
00:08:03,560 --> 00:08:06,710
each coefficient must be 0.
116
00:08:06,710 --> 00:08:10,330
Therefore, we have
that j plus 2,
117
00:08:10,330 --> 00:08:28,810
times j plus 1, times aj plus
2, is equal to 2 j plus 1 minus
118
00:08:28,810 --> 00:08:32,520
e, aj.
119
00:08:32,520 --> 00:08:38,280
I set this whole combination
inside brackets to 0.
120
00:08:38,280 --> 00:08:43,710
So this term is equal to
this term and that term
121
00:08:43,710 --> 00:08:44,970
on the other side.
122
00:08:44,970 --> 00:08:47,330
You get a plus 2j.
123
00:08:47,330 --> 00:08:51,210
A plus 1 and minus e.
124
00:08:51,210 --> 00:08:56,165
So basically this is
a recursion relation.
125
00:08:56,165 --> 00:09:03,855
Aj plus 2 is equal
to 2j plus 1 minus e,
126
00:09:03,855 --> 00:09:10,050
over j plus 2, j
plus 1 times aj.
127
00:09:17,370 --> 00:09:19,870
And this is perfectly nice.
128
00:09:19,870 --> 00:09:22,630
This is what should
have happened
129
00:09:22,630 --> 00:09:26,200
for this kind of
differential equation--
130
00:09:26,200 --> 00:09:31,820
a second-order linear
differential equation.
131
00:09:31,820 --> 00:09:35,320
We get a recursion
that jumps one step.
132
00:09:35,320 --> 00:09:37,205
That's very nice.
133
00:09:37,205 --> 00:09:42,700
And this should hold
for j equals 0, 1, 2--
134
00:09:42,700 --> 00:09:43,420
all numbers.
135
00:09:43,420 --> 00:09:47,960
So when you start solving this,
there's two ways to solve it.
136
00:09:47,960 --> 00:09:52,977
You can decide, OK, let me
assume that you know a0--
137
00:09:52,977 --> 00:09:54,290
you give it.
138
00:09:54,290 --> 00:09:56,500
Give a0.
139
00:09:56,500 --> 00:10:00,215
Well, from this equation--
from a0-- you can calculate a2.
140
00:10:00,215 --> 00:10:03,070
And then from a2 you
can calculate a4.
141
00:10:03,070 --> 00:10:04,745
And successively.
142
00:10:04,745 --> 00:10:09,160
So you get a2, a4--
143
00:10:09,160 --> 00:10:09,940
all of those.
144
00:10:09,940 --> 00:10:17,350
And this corresponds to an even
solution of the differential
145
00:10:17,350 --> 00:10:21,010
equation for h.
146
00:10:21,010 --> 00:10:24,010
Even coefficients.
147
00:10:24,010 --> 00:10:26,290
Even solution for h.
148
00:10:29,630 --> 00:10:36,190
Or-- given this recursion--
you could also give a1--
149
00:10:36,190 --> 00:10:41,210
give it-- and then
calculate a3, a5--
150
00:10:41,210 --> 00:10:43,354
and those would be
an odd solution.
151
00:10:47,310 --> 00:10:51,375
So you need two
conditions to solve this.
152
00:10:54,420 --> 00:11:01,990
And those conditions
are a0 and a1,
153
00:11:01,990 --> 00:11:09,570
which is the same as specifying
the value of the function h
154
00:11:09,570 --> 00:11:12,580
at 0--
155
00:11:12,580 --> 00:11:17,970
because the value of the
function h at 0 is a0.
156
00:11:17,970 --> 00:11:21,460
And the value of the derivative
of the function at 0,
157
00:11:21,460 --> 00:11:22,593
which is a1.
158
00:11:26,537 --> 00:11:36,910
[INAUDIBLE] h of mu is a0
plus a1u plus a2u squared.
159
00:11:36,910 --> 00:11:40,040
So the derivative
at 0 is [? h1. ?]
160
00:11:40,040 --> 00:11:42,730
And that's what you must have
for solving a differential
161
00:11:42,730 --> 00:11:46,990
equation-- a second-order
differential equation for h.
162
00:11:46,990 --> 00:11:51,010
You need to know the value
of the function at zero
163
00:11:51,010 --> 00:11:54,920
and the value of the
function at the derivative
164
00:11:54,920 --> 00:11:56,500
of the function--
165
00:11:56,500 --> 00:11:58,110
at zero, as well.
166
00:11:58,110 --> 00:12:00,310
And then you can
start integrating it.
167
00:12:00,310 --> 00:12:08,670
So this first gives you a
solution a0 plus a2u squared,
168
00:12:08,670 --> 00:12:12,780
plus a4u to the fourth.
169
00:12:12,780 --> 00:12:21,640
And the second is an a1u plus
a3u cubed, plus these ones.
170
00:12:21,640 --> 00:12:26,220
So all looks pretty much OK.